Problem: An online retailer offers next-day shipping for an extra fee. The retailer says that $95\%$ of customers who pay for next-day shipping actually receive the item the next day, and those who don't are issued a refund. Suppose we take an SRS of $20$ next-day orders, and let $X$ represent the number of these orders that arrive the next day. Assume that the arrival statuses of orders are independent. Which of the following would find $P(X=19)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A ${20 \choose 19}(0.95)(0.05)^{19}$ (Choice B) B ${20 \choose 19}(0.95)^{19}(0.05)$ (Choice C) C ${95 \choose 20}(0.95)^{19}(0.05)$ (Choice D) D $(0.95)(0.05)^{19}$ (Choice E) E $(0.95)^{19}(0.05)$
Answer: Probability of $19$ successes We want the probability that there are $19$ successes (next day arrivals) in $20$ trials (number of orders), so we're going to need $1$ failure (late arrival) as well. The probability of each success is ${0.95}$ and the probability of each failure is $0.05}$. Since we were told to assume independence, we can multiply probabilities to find the probability of getting $19$ successes followed by $1$ failure: $\begin{aligned} P(\text{SS}\dots\text{SF})&=\left({0.95}\right)\left({0.95}\right)\dots\left({0.95}\right)\left(0.05}\right) \\\\ &=\left({0.95}\right)^{19}\left(0.05}\right) \end{aligned}$ The binomial coefficient ${n \choose k}$ Nineteen successes followed by $1$ failure isn't the only arrangement that produces $19$ successes in $20$ trials. For instance, the single failure could occur anywhere in the sequence of $20$ trials and still produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=19$ successes (next day arrivals) in $n=20$ trials (number of orders), so we should use the binomial coefficient ${20 \choose 19}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $(0.95)^{19}(0.05)$ so for our final answer we multiply this probability by the number of possible arrangements: ${20 \choose 19}(0.95)^{19}(0.05)$ The answer: ${20 \choose 19}(0.95)^{19}(0.05)$